Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (Fig. P5.43). Suppose F = 68.0 N, m1 = 12.0 kg, m2 = 18.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.100. (a) Draw a free-body diagram for each block. (b) Determine the tension T and the magnitude of the acceleration of the system.
F = ma - fk
fk = μkmg
The Attempt at a Solution
68.0 N = (12.0 kg + 18.0 kg)*a – (0.100)(12.0 kg)(9.8 m/s2) – (0.100)(18.0 kg)(9.8 m/s2)
a = [68.0 N + (0.100)(12.0 kg)(9.8 m/s2) + (0.100)(18.0 kg)(9.8 m/s2)]/(12.0 kg + 18.0 kg)
a = 3.25 m/s2
T = ma - fk
T = (12.0 kg)( 3.25 m/s2) - (0.100)(12.0 kg)(9.8 m/s2)
T = 27.2 N
The tension of T is correct according to the answer key in the book but my acceleration is not. I cannot understand this since the acceleration I have is what I used to determine T. The book states a = 1.29 m/s2
The only thing I can think of is that since friction is involved that there must be a vertical acceleration that must be taken into account then I could square each acceleration of the x and y direction, add them and take the square root. Am I on the right track and how would I determine the acceleration in the y direction?